Step 1: Identify ions in aqueous NaCl solution.
In aqueous NaCl: $Na^+$, $Cl^-$, $H^+$ (from water), $OH^-$ (from water) are all present. During electrolysis, cations ($Na^+$, $H^+$) migrate to cathode; anions ($Cl^-$, $OH^-$) migrate to anode.
Step 2: Discharge at cathode.
At the cathode (reduction), $H^+$ is preferred over $Na^+$ (Na deposition from aqueous solution requires very high energy). So: \[ 2H^+ + 2e^- \rightarrow H_2 \uparrow \] $H_2$ is always evolved at cathode in aqueous NaCl.
Step 3: Discharge at anode in concentrated NaCl.
When $Cl^-$ concentration is high, $Cl^-$ is preferentially oxidised: $2Cl^- \rightarrow Cl_2 + 2e^-$. Chlorine gas evolves at anode in concentrated NaCl.
Step 4: Discharge at anode in very dilute NaCl (the key distinction).
In very dilute NaCl, $Cl^-$ concentration is very low. Water is preferentially oxidised at the anode: \[ 2H_2O \rightarrow O_2 \uparrow + 4H^+ + 4e^- \] $O_2$ gas is evolved at the anode in very dilute NaCl.
Step 5: Evaluate the options.
Option 1: $H_2$ at anode - Wrong ($H_2$ is at cathode). Option 2: Na at cathode - Wrong ($H_2$ is produced at cathode). Option 3: $O_2$ at anode - Correct! Specific to very dilute NaCl. Option 4: $H_2$ at cathode - Although true, this is not specific to dilute NaCl (it happens always). The most distinctive correct statement for very dilute NaCl is option 3.
Step 6: State the final answer.
In electrolysis of very dilute aqueous NaCl, $O_2$ gas is evolved at the anode (water is oxidised). \[ \boxed{O_2 \text{ gas is evolved at the anode}} \]